Finding a basis for a field extension $\mathbb{Q}(i,\sqrt[4]{2})/\mathbb{Q}(\sqrt{2}).$

Question

I am looking to find a basis for the field extension $$\mathbb{Q}(i,\sqrt[4]{2})/\mathbb{Q}(\sqrt{2}).$$

Clearly as far as I can tell the elements $i$ and $\sqrt[4]{2}$ would be in the basis, as neither of these elements is in the field $\mathbb{Q}(\sqrt{2})$. However, when it comes to finding other elements in the basis, all that I could really think of at first was $i\sqrt[4]{2}$. Then I supposed that we would also need $\sqrt[4]{2}^3$ and $i\sqrt[4]{2}^3$ in the basis too, and later wondered whether $i \sqrt{2}$ would also be required. I think not because it would be covered under the element $i$ as $\sqrt{2}$ is contained in the field but I am not certain.

As it took me a while to find each of these 5 elements, I am wondering whether this is actually all the elements that would be required to form a basis, whether I am missing any elements or have too many. Any help would be appreciated thanks :)

Answer 1

Note that $i\notin\mathbb{Q}(\sqrt[4]{2})$, so $\mathbb{Q}(i,\sqrt[4]{2})$ has degree $8$ over $\mathbb{Q}$. Hence it has degree $4$ over $\mathbb{Q}(\sqrt{2})$.

A basis for $\mathbb{Q}(\sqrt[4]{2})$ over $\mathbb{Q}(\sqrt{2})$ is $\{1,\sqrt[4]{2})$.

Can you finish? You still have to add $i$.

Answer 2

I think (2^¼)^2=√2 this imply that √2∈Q(2^¼) that is Q(√2) Contained in Q(2^¼) therefore degree of i over Q(√2) is 2 that is degree of Q(2^¼, i) over Q(√2) is 2 therefore basis of Q(2^¼, i) over Q(√2) contains only two element namely {1,i}.